You're correct in your approach! Let's break down the process clearly:

Given the expression:

$\frac{2\sqrt{10}}{\sqrt{7} - \sqrt{5}}$

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator:

$\frac{2\sqrt{10}}{\sqrt{7} - \sqrt{5}} \times \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} + \sqrt{5}} = \frac{2\sqrt{10}(\sqrt{7} + \sqrt{5})}{(\sqrt{7})^2 - (\sqrt{5})^2}$

Now, simplify the denominator:

$(\sqrt{7})^2 - (\sqrt{5})^2 = 7 - 5 = 2$

So the expression becomes:

$\frac{2\sqrt{10}(\sqrt{7} + \sqrt{5})}{2}$

Now, distribute $2\sqrt{10}$ in the numerator:

$\frac{2\sqrt{10}\sqrt{7} + 2\sqrt{10}\sqrt{5}}{2} = \frac{2\sqrt{70} + 2\sqrt{50}}{2}$

Finally, simplify by dividing the numerator by 2:

$\sqrt{70} + \sqrt{50}$

But we know that $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$. Therefore:

$\sqrt{70} + 5\sqrt{2}$

So the final simplified expression is:

$\sqrt{70} + 5\sqrt{2}$

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Do you want further details or have any questions?

Here are 5 questions that expand on the above concepts:

1. Can you rationalize another expression similar to this one?
2. Why is the conjugate useful in simplifying radicals?
3. How do you handle more complex radical expressions in the numerator?
4. What happens if the denominator is a binomial with a sum instead of a difference?
5. Can you simplify an expression involving cube roots using a similar approach?

Tip: Always look for opportunities to simplify the radicals themselves, as this can often lead to a more straightforward expression.